finding another particular solution of linear ODE

Consider thehomogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation)second-order linear ordinary differential equation

$y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=\mathrm{\hspace{0.33em}0}.$(1)

If one knows oneparticular solution (http://planetmath.org/SolutionsOfOrdinaryDifferentialEquation) $y=y_1(x)\not\equiv 0$  of (1), it’s possible to derivefrom it via two quadratures another solution  $y_2(x)$, linearly independent on  $y_1(x)$;  thus one can write thegeneral solution

$$y=C_1{y}_1(x)+C_2{y}_2(x)$$

of that homogeneous differential equation.

We will now show the derivation procedure.

We put

$y=uv$(2)

which renders (1) to

$(v^{\prime \prime }+P{v}^{\prime }+Qv)u+(2v^{\prime }+Pv)u^{\prime }+u^{\prime \prime }v=\mathrm{\hspace{0.33em}0}.$(3)

Here one can choose  $v:=y_1(x)$, whence the first addendvanishes, and (3) gets the form

$(2y_1^{\prime }+P{y}_1)u^{\prime }+y_1{u}^{\prime \prime }=\mathrm{\hspace{0.33em}0}.$(4)

This equation may be written as  $\frac{u^{\prime \prime }}{u^{\prime }}=-2\frac{y_1^{\prime }}{y_1}-P$, which is integrated to

$$\ln \left|\frac{du}{dx}\right|=\ln \frac{1}{y_1^2}-\int P𝑑x+\text{constant},$$

i.e.

$$\frac{du}{dx}=\frac{C}{y_1^2}{e}^{-\int P𝑑x}.$$

A new integration results from this the general solution of (4):

$$u=C\int \frac{e^{-\int P𝑑x}}{y_1^2}𝑑x+C^{\prime }.$$

Thus by (2), we have obtained the wanted other solution

$$y_2(x)=y_1(x)\int \frac{e^{-\int P𝑑x}}{y_1^2}𝑑x$$

which is clearly linearly independent on y_1(x).

Consequently, we can express the general solution of thedifferential equation (1) as

$$y=y_1(x)u=C_1{y}_1(x)+C_2{y}_1(x)\int \frac{e^{-\int P𝑑x}}{y_1^2}𝑑x,$$

where $C_1$ and $C_2$ are arbitrary constants.

Remark.  The substitution

$$y:=e^{-\frac{1}{2}\int P(x)𝑑x}u$$

converts the equation (1) into the form

$$\frac{d^{2}u}{d{x}^2}+(Q-\frac{P^2}{4}-\frac{P^{\prime }}{2})u=\mathrm{\hspace{0.33em}0}$$

not containing the derivative $\frac{du}{dx}$.